Möbius band diffrential geometry

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hi, Initially, I know how to take surface integral of möbius band via given parameterization, but I really wonder how these parameters are created. How can we derive these parameters ?? What is the logic of deriving such a good parameters?? Could you give some proofs??
 

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  • #2
Orodruin
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The surface integral over a Möbius strip is not well defined. Since the area form needs to be oriented, you cannot define it properly on a non-orientable manifold. You can only perform the integral if you do not care about the orientation of the area form. The Möbius strip itself as a smooth manifold is defined by an atlas.
 
  • #3
stevendaryl
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This is a technical point that doesn't get a lot of attention--presumably because it doesn't come up that often. It's discussed by John Baez in an old sci.physics.research Usenet group. https://groups.google.com/forum/#!topic/sci.physics.research/aiMUJrOjE8A[151-175]

What Baez says, if I'm paraphrasing correctly, is that to integrate over a manifold, what you need is not a form, but a pseudo-form. Pseudo-forms are analogous to pseudo-scalars. A pseudo-scalar behaves like a scalar, for most purposes, but acquires an extra minus sign under a parity inversion.

Here's what I think is the simplest example illustrating the point:

Suppose we have a little section of a 1-D manifold, and we have a scalar function [itex]\phi[/itex], and we'd like to integrate [itex]\phi[/itex] over that manifold. Well, to get a number out of it, we have to choose a coordinate system, which for 1-D might as well be the coordinate [itex]x[/itex]. So we pick a coordinate system in which [itex]x[/itex] goes from 0 to 1, and we can integrate as follows:

[itex]I = \int_0^1 \phi(x) dx[/itex]

But what if you had instead chosen a different coordinate system [itex]y = -x[/itex]? Then the integral [itex]\int_0^1 \phi(y) dy[/itex] would given the negative result. So to get a unique answer that is coordinate-independent, instead of using a form [itex]dx[/itex], you should use a pseudo-form [itex]|dx|[/itex].

You can integrate a pseudo-form over a non-orientable manifold.

To quote Wikipedia here: https://en.wikipedia.org/wiki/Differential_form

On an orientable but not oriented manifold, there are two choices of orientation; either choice allows one to integrate n-forms over compact subsets, with the two choices differing by a sign. On non-orientable manifold, n-forms and densities cannot be identified —notably, any top-dimensional form must vanish somewhere (there are no volume forms on non-orientable manifolds), but there are nowhere-vanishing densities— thus while one can integrate densities over compact subsets, one cannot integrate n-forms. One can instead identify densities with top-dimensional pseudoforms.
 
  • #4
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This is a technical point that doesn't get a lot of attention--presumably because it doesn't come up that often. It's discussed by John Baez in an old sci.physics.research Usenet group. https://groups.google.com/forum/#!topic/sci.physics.research/aiMUJrOjE8A[151-175]

What Baez says, if I'm paraphrasing correctly, is that to integrate over a manifold, what you need is not a form, but a pseudo-form. Pseudo-forms are analogous to pseudo-scalars. A pseudo-scalar behaves like a scalar, for most purposes, but acquires an extra minus sign under a parity inversion.

Here's what I think is the simplest example illustrating the point:

Suppose we have a little section of a 1-D manifold, and we have a scalar function [itex]\phi[/itex], and we'd like to integrate [itex]\phi[/itex] over that manifold. Well, to get a number out of it, we have to choose a coordinate system, which for 1-D might as well be the coordinate [itex]x[/itex]. So we pick a coordinate system in which [itex]x[/itex] goes from 0 to 1, and we can integrate as follows:

[itex]I = \int_0^1 \phi(x) dx[/itex]

But what if you had instead chosen a different coordinate system [itex]y = -x[/itex]? Then the integral [itex]\int_0^1 \phi(y) dy[/itex] would given the negative result. So to get a unique answer that is coordinate-independent, instead of using a form [itex]dx[/itex], you should use a pseudo-form [itex]|dx|[/itex].

You can integrate a pseudo-form over a non-orientable manifold.

To quote Wikipedia here: https://en.wikipedia.org/wiki/Differential_form
I scrutinized, and looked very deeply at the parametrisation of möbius band, but, I recognised that all z value of coordinate system just depended on COS(t/2) in my attachments. So, my question: Are we sure that this z coordinate value definitely depends on COS(t/2) ????, Can't we have COS (t/3), COS (t/4).... ???, Is there a proof that only COS(t/2) satisfies mobius band??? (( I am aware that COS(t/2) make a sense to satisfy the x,y,z values at t=0 or t=2 \pi according to my attachments, besides all my questions may turn out: Can't we have COS (t/3) or COS (t/4) .... which satisfies the x,y,z values at t=0 or t=2 \pi and affect the all z value of coordinate???????? )))
 

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hi, I have seen that I can not receive an answer for a long time for my last question. Did I express my question in a complicated form????? or Can't you understand the question because of the situation that I mixed it up????. Besides, I will be really delighted if you return to me.....
 
  • #6
stevendaryl
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hi, I have seen that I can not receive an answer for a long time for my last question. Did I express my question in a complicated form????? or Can't you understand the question because of the situation that I mixed it up????. Besides, I will be really delighted if you return to me.....
I'm sorry, but I don't quite follow what you're doing. I assume that you are embedding the mobius strip in 3D, but I'm not sure I understand your embedding. According to Wikipedia, a common embedding is:

[itex]x = (1+ \frac{v}{2} cos(\frac{u}{2})) cos(u)[/itex]
[itex]y = (1+\frac{v}{2} cos(\frac{u}{2})) sin(u)[/itex]
[itex]z = \frac{v}{2} sin(\frac{u}{2})[/itex]
 
  • #7
stevendaryl
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I'm sorry, but I don't quite follow what you're doing. I assume that you are embedding the mobius strip in 3D, but I'm not sure I understand your embedding. According to Wikipedia, a common embedding is:

[itex]x = (1+ \frac{v}{2} cos(\frac{u}{2})) cos(u)[/itex]
[itex]y = (1+\frac{v}{2} cos(\frac{u}{2})) sin(u)[/itex]
[itex]z = \frac{v}{2} sin(\frac{u}{2})[/itex]
This embedding is the surface generated by a line segment of length [itex]v[/itex] that moves around a circle of radius [itex]1[/itex], rotating as it goes.
 
  • #8
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This embedding is the surface generated by a line segment of length [itex]v[/itex] that moves around a circle of radius [itex]1[/itex], rotating as it goes.
Initially, I understand the logic of parametrisation you shared, "u" is the rotating angle in "xy plane", and "u/2" is the angle between "z" and "xy" plane, and my question is: Do we have to have "u/2" angle to indicate the x,y,z??? Can't it be "u/3" instead of "u/2"??? ( As you can see x,y,z values always involves the "u/2") Why can't it be another angle???
 
  • #9
stevendaryl
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Initially, I understand the logic of parametrisation you shared, "u" is the rotating angle in "xy plane", and "u/2" is the angle between "z" and "xy" plane, and my question is: Do we have to have "u/2" angle to indicate the x,y,z??? Can't it be "u/3" instead of "u/2"??? ( As you can see x,y,z values always involves the "u/2") Why can't it be another angle???
Think of constructing a Mobius strip this way:
  • Take a wire of length [itex]2 \pi[/itex], and bend it into a circle. Let [itex]u[/itex] measure the distance around the circle.
  • Take a bunch of matchsticks, and place them around the circle, with the center of each match touching the wire, and with the match running perpendicular to the wire. Glue the match to the circle at the center, after choosing the appropriate orientation (described below).
  • At the point [itex]u=0[/itex], orient a matchstick with the head pointing radially away from the center.
  • At the point [itex]u=\pi[/itex], orient a matchstick with the head pointing straight up.
  • At the point [itex]u= 2 \pi[/itex], orient a matchstick with the head pointing radially in, toward the center. (This match is right next to the starting one, but is pointing in the opposite direction).
  • At intermediate points, rotate the matches so that they smoothly interpolate between the three described above.
  • To complete the Mobius strip, take a rectangular strip of paper with length [itex]2 \pi[/itex] and width the same as the matchsticks. Glue one edge of the strip to the heads of the matches, and the other edge to the tails.
So [itex]u[/itex] measures how far around the circle you've gone, and it runs from [itex]0[/itex] to [itex]2 \pi[/itex]. But [itex]\frac{u}{2}[/itex] measures how much the matches have rotated. Going all the way around the circle causes the matches to rotate to the opposite direction, which means they have rotated through [itex]\pi[/itex] radians, not [itex]2 \pi[/itex].
 
  • #10
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Think of constructing a Mobius strip this way:
  • Take a wire of length [itex]2 \pi[/itex], and bend it into a circle. Let [itex]u[/itex] measure the distance around the circle.
  • Take a bunch of matchsticks, and place them around the circle, with the center of each match touching the wire, and with the match running perpendicular to the wire. Glue the match to the circle at the center, after choosing the appropriate orientation (described below).
  • At the point [itex]u=0[/itex], orient a matchstick with the head pointing radially away from the center.
  • At the point [itex]u=\pi[/itex], orient a matchstick with the head pointing straight up.
  • At the point [itex]u= 2 \pi[/itex], orient a matchstick with the head pointing radially in, toward the center. (This match is right next to the starting one, but is pointing in the opposite direction).
  • At intermediate points, rotate the matches so that they smoothly interpolate between the three described above.
  • To complete the Mobius strip, take a rectangular strip of paper with length [itex]2 \pi[/itex] and width the same as the matchsticks. Glue one edge of the strip to the heads of the matches, and the other edge to the tails.
So [itex]u[/itex] measures how far around the circle you've gone, and it runs from [itex]0[/itex] to [itex]2 \pi[/itex]. But [itex]\frac{u}{2}[/itex] measures how much the matches have rotated. Going all the way around the circle causes the matches to rotate to the opposite direction, which means they have rotated through [itex]\pi[/itex] radians, not [itex]2 \pi[/itex].
Initially, thank you for your return. Nevertheless, as your reply, we can give examples for just these three points(0,pi,2*pi) in order to show that "u/2" measures how much the matches have rotated. I admit that "u/2" is so suitable for these three points, but How do we know that "u/2" angle satisfies all intermediate points between 0 and pi, or between pi and 2*pi ????? Because I consider that saying that "u/2" angle definitely suits all points by just looking that "u/2" suits the points (0,pi,2*pi) very well is misleading....Am I right??
 
  • #11
stevendaryl
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Initially, thank you for your return. Nevertheless, as your reply, we can give examples for just these three points(0,pi,2*pi) in order to show that "u/2" measures how much the matches have rotated. I admit that "u/2" is so suitable for these three points, but How do we know that "u/2" angle satisfies all intermediate points between 0 and pi, or between pi and 2*pi ????? Because I consider that saying that "u/2" angle definitely suits all points by just looking that "u/2" suits the points (0,pi,2*pi) very well is misleading....Am I right??
What angle for an intermediate value of [itex]u[/itex] would be "suitable"? What do you mean by "suitable"?

A mobius strip is a topological object. It does not have a specific shape, in that you can twist and bend and stretch it and it will still be a mobius strip. The only constraint is that from one end of the strip to the other, the strip makes a 180 degree rotation (or [itex]2 \pi[/itex]) radians. How much it twists along the way is irrelevant, but the parametrization I gave distributes the "twist" evenly along the strip.
 
  • #12
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I mean " Do we have other options to satisfy the intermediate points besides the u/2 ??? ( why can't we have u/3,u/4... ???)
 
  • #13
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Also, I consider that saying that "u/2" angle definitely suits all points by just admitting that "u/2" suits the points (0,pi,2*pi) very well is misleading.
 
  • #14
stevendaryl
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I mean " Do we have other options to satisfy the intermediate points besides the u/2 ??? ( why can't we have u/3,u/4... ???)
If you use [itex]u/3[/itex], then the strip of paper will only be twisted by 120 degrees (or [itex]\frac{2}{3} \pi[/itex] radians) when you go all the way around. So the two ends of the strip will not be parallel, so you can't glue them together to make a Mobius strip.
 
  • #15
stevendaryl
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Also, I consider that saying that "u/2" angle definitely suits all points by just admitting that "u/2" suits the points (0,pi,2*pi) very well is misleading.
No, it's not at all misleading. There are two "angles" involved: [itex]u[/itex], which measures how far around the circle you've gone, and--let me call the second angle [itex]\alpha[/itex], which measures how much the strip has been twisted, compared to the start.

You need to pick [itex]\alpha = f(u)[/itex], with the only constraints being:
  • [itex]f(2 \pi) = \pi + f(0)[/itex]
  • [itex]f(u)[/itex] is continuous for [itex]0 < u < 2\pi[/itex]
  • Probably you also want [itex]\frac{df}{du}[/itex] to be continuous, as well.
  • And probably you also want [itex]\frac{df}{du}|_{u=0} = \frac{df}{du}|_{u=2\pi}[/itex], also.
The choice [itex]f(u) = u/2[/itex] is the simplest choice, but you could have other functions. But you can't have [itex]f(u) = \frac{u}{3}[/itex]
 
  • #16
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No, it's not at all misleading. There are two "angles" involved: [itex]u[/itex], which measures how far around the circle you've gone, and--let me call the second angle [itex]\alpha[/itex], which measures how much the strip has been twisted, compared to the start.

You need to pick [itex]\alpha = f(u)[/itex], with the only constraints being:
  • [itex]f(2 \pi) = \pi + f(0)[/itex]
  • [itex]f(u)[/itex] is continuous for [itex]0 < u < 2\pi[/itex]
  • Probably you also want [itex]\frac{df}{du}[/itex] to be continuous, as well.
  • And probably you also want [itex]\frac{df}{du}|_{u=0} = \frac{df}{du}|_{u=2\pi}[/itex], also.
The choice [itex]f(u) = u/2[/itex] is the simplest choice, but you could have other functions. But you can't have [itex]f(u) = \frac{u}{3}[/itex]
Thank you "stevendarly", Final question ( I am asking because of my curious :D ). Could you give another parameterization of möbius band????? ( Besides the parameterization we can see everywhere)
 
  • #17
Orodruin
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Thank you "stevendarly", Final question ( I am asking because of my curious :D ). Could you give another parameterization of möbius band????? ( Besides the parameterization we can see everywhere)
What you just discussed was not a parametrisation, it was an embedding. In order to define the Möbius strip with an atlas you are going to need at least two charts. Of course, you can then introduce other coordinate systems at your leisure.

The embedding in ##\mathbb R^3## is not a property of the Möbius strip itself and there is an infinite number of possibilities. Just like there are infinite possibilities of embedding the sphere.
 

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